This video will give you an idea of how powerful a falling/rolling rubberband ball can be.
2006-07-28 15:52:07
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answer #1
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answered by PC_Load_Letter 4
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The main factor is the properties of the rubber used in the elastic ball.
Super rubber was developed some years ago and resembled a soft plastic which could be rolled into a ball in the hand, much like dough.
This material gave a far greater 'elasticity' to the ball which consequently bounced higher and rebounded more times than a 'normal' elastic ball.
The height of the drop IS a factor influencing the NUMBER of times the ball would rebound. To prove this drop the ball from a height of 2 cm onto a hard surface and count the rebounds.
Repeat from a height of 2 m and count the rebounds.
In general terms the ball has kinetic energy when it strikes the ground. In coming to rest the ball commpresses and flattens against the surface.
It now has potential energy and acts precisely as a spring. It releases the stored potential energy which causes the ball to rebound. In compressing, and stretching, some heat energy is generated and thus the kinetic energy is reduced, Eventually the ball comes to rest when all the energy is dissipated.
A thin wire coiled into a fairly large diameter helix with a small weight at the end will stretch and rebound a large number of times. (See toys in toy shops)
A thick wire coiled into a fairly small diameter helix with a similar weight at the end will stretch to a much smaller degree and rebound only a very few times and may not even show any rebound at all.
To be more technical, materials have a number of properties and the elasticity of a material is related to 'Young's Modulus of Elasticity' for that material.
Without this information the question cannot be resolved.
As a matter of interest glass also possess the property of elasticity. Elastic glass? Yes, drop a glass marble onto a hardened surface and count how many times it rebounds!
2006-07-28 01:22:09
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answer #2
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answered by CurlyQ 4
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Very confusing words but I think you intend to drop a bouncy ball off a very high building and figure what happens.?
As it goes down first time it accelerates under the force of gravity to its 'terminal velocity in air'. maybe about 120 mph. No matter how far it falls it will not exceed that speed. It reaches (T.V.air) about 5 secs after release and maybe 500 feet down.
When it hits the ground it flattens against the syrface and collects energy because of its elastic properties and bounces up again. But on the way up gravity is working against it. Slowing it down until it reaches speed zero. That point is VERY hard to determine.
But remember after 500 feet high it does not matter what height the building is? Lets decide to make the building 500 feet high?
So it might climb about 250 feet before it starts to fall for the second time. When it hits the ground for the 2nd time it will be travelling slower than 120mph. So the next up bounce will be LESS than half its falling height.
It will continue to bounce less and less every bounce until eventually it is rising so little that gravity takes over completely and it comes to rest.
NOW FOR THE ANSWER
Believe it or not: If you take a small bouncy ball and drop it from head height the number of bounces, [in real life] will be much the same as if you use the tallest building on earth!
Brilliant question my friend and thank you for giving me a chance to resurrect my tired old brain
2006-07-27 23:53:41
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answer #3
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answered by SouthOckendon 5
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You definitiely need to know the properties of the properties of the elastic more specifically the modulus of elasticity.
The modulus of elasticity is the a figure between 0 and 1. At zero there is no "elastic collision" (the ball doesn't bounce) and at 1 it is a perfect elastic collision. That is if you drop an elastic ball from 2m it will bounce back up 2m.
The modulus of elsticity can be found by dividing the relative velocity after the bounce by the relative velocity before the bounce. This equation can be rearranged to give the equation
Vel after bounce = Mod of Elasticity x Vel before bounce.
As there is never a perfectly elastic collision the modulus of elasticity is always less than one so each and every bounce you are multiplying the velocity by a figure less than 1 and the vel after the bounce decreases until it eventually reaches zero. You can use this idea to further develop your equation as follows
Vel after bounce = Mod of elasticity x (Mod of elasticity x ... (Mod of elasticity x Vel before bounce)))))
This can be tidied up to
Vel after bounce = (Mod of Elasticity^n) x Vel after bounce
Where n is the number of times the ball bounces.
Provided you know the Vel before the bounce, which will increase with height until you get the terminal velocity as mentioned in other posts, you can solve this equation by then substuting in zero as the Vel after the bounce, as the ball has come to rest.
E.g.
0 = (Mod of elasticity^n) x Vel before bounce.
You can now apply this to any bouncing object you wish!
Enjoy
2006-07-28 01:46:44
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answer #4
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answered by dope_move_busta 1
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the ball is stopping when the hieght is zero , if the hieght of building is H the hieght is increased each ball bounce by the amount vt-1/2(gt^2)n where v inital velocity g gravity accleration , t time of hieght, n number of ball bounce , t =(v/g) so that when the ball rest on ground H=0 then v(v/2g)- (1/2)ng(v/2g)^2=0 or v^2/2g=1/2(nv^2)(1/4g) solving for n , n= 4 times ( regardless of the hieght value) ,
2006-07-28 09:25:26
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answer #5
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answered by abduasslamalgattawi 2
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Actually, someone tried this once, only it was a superball. They dropped the world's biggest bouncy ball off the sears tower, only to have it shatter and smash things all around the city.
So, it wouldn't bounce back up, it'd just break.
2006-07-28 02:19:15
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answer #6
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answered by themuffinking01 2
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S= size of elastcic band ball
H= high of building
J= jump number
E= high of jump J
IF H=>501M H MUST CHANGE IT SELT TO 500M TO WORK
IF S=>251M H MUST CHANGE IT SELT TO 250M TO WORK
H*S+S-H-((j- 1)/2)=e
EG
500*250+250-500-(1-1)/2=124750
500*250+250-500-(2-1)/2=124749.5
THEN
124749
124748.5
124748
124747.5
124747
124746.5
124746
124745.5
124745
124744.5
124744
124743.5
2006-07-28 00:24:54
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answer #7
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answered by green_maths_scout 2
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i'd guess it would bounce as many times as any other size elastic band ball
2006-07-27 23:51:03
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answer #8
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answered by Anonymous
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It would destroy itself on impact as the bands being stretched around it would not have any elasticity to allow it to bounce.
2006-07-27 23:35:46
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answer #9
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answered by rogue_samurai 3
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ER??? Odd question, but interesting, sorry I don't know the answer. Is this a trick question??
2006-07-28 02:00:58
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answer #10
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answered by Jensen Ackles Girl (I Wish!) 5
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